Difference between revisions of "2019 AMC 10B Problems/Problem 6"
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− | There is a | + | {{duplicate|[[2019 AMC 10B Problems|2019 AMC 10B #6]] and [[2019 AMC 12B Problems|2019 AMC 12B #4]]}} |
+ | |||
+ | ==Problem== | ||
+ | |||
+ | There is a positive integer <math>n</math> such that <math>(n+1)! + (n+2)! = n! \cdot 440</math>. What is the sum of the digits of <math>n</math>? | ||
<math>\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12</math> | <math>\textbf{(A) }3\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad\textbf{(E) }12</math> | ||
==Solution== | ==Solution== | ||
− | < | + | ===Solution 1=== |
− | <math>n! | + | |
− | <math>n + 1 + n | + | <cmath>\begin{split}& (n+1)n! + (n+2)(n+1)n! = 440 \cdot n! \\ |
− | <math>n | + | \Rightarrow \ &n![n+1 + (n+2)(n+1)] = 440 \cdot n! \\ |
+ | \Rightarrow \ &n + 1 + n^2 + 3n + 2 = 440 \\ | ||
+ | \Rightarrow \ &n^2 + 4n - 437 = 0\end{split}</cmath> | ||
+ | |||
+ | Solving by the quadratic formula, <math>n = \frac{-4\pm \sqrt{16+437\cdot4}}{2} = \frac{-4\pm 42}{2} = \frac{38}{2} = 19</math> (since clearly <math>n \geq 0</math>). The answer is therefore <math>1 + 9 = \boxed{\textbf{(C) }10}</math>. | ||
+ | |||
+ | ===Solution 2=== | ||
+ | |||
+ | Dividing both sides by <math>n!</math> gives | ||
+ | <cmath>(n+1)+(n+2)(n+1)=440 \Rightarrow n^2+4n-437=0 \Rightarrow (n-19)(n+23)=0.</cmath> | ||
+ | Since <math>n</math> is non-negative, <math>n=19</math>. The answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | |||
+ | Dividing both sides by <math>n!</math> as before gives <math>(n+1)+(n+1)(n+2)=440</math>. Now factor out <math>(n+1)</math>, giving <math>(n+1)(n+3)=440</math>. By considering the prime factorization of <math>440</math>, a bit of experimentation gives us <math>n+1=20</math> and <math>n+3=22</math>, so <math>n=19</math>, so the answer is <math>1 + 9 = \boxed{\textbf{(C) }10}</math>. | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/ba6w1OhXqOQ?t=1956 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/7xf_g3YQk00 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | https://youtu.be/6YFN_hwotUk | ||
+ | |||
+ | ~savannahsolver | ||
− | + | ==See Also== | |
− | + | {{AMC10 box|year=2019|ab=B|num-b=5|num-a=7}} | |
+ | {{AMC12 box|year=2019|ab=B|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Revision as of 04:26, 15 January 2021
- The following problem is from both the 2019 AMC 10B #6 and 2019 AMC 12B #4, so both problems redirect to this page.
Contents
Problem
There is a positive integer such that . What is the sum of the digits of ?
Solution
Solution 1
Solving by the quadratic formula, (since clearly ). The answer is therefore .
Solution 2
Dividing both sides by gives Since is non-negative, . The answer is .
Solution 3
Dividing both sides by as before gives . Now factor out , giving . By considering the prime factorization of , a bit of experimentation gives us and , so , so the answer is .
Video Solution
https://youtu.be/ba6w1OhXqOQ?t=1956
~ pi_is_3.14
Video Solution
~IceMatrix
~savannahsolver
See Also
2019 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 3 |
Followed by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.